GepGreeloCesyjk

## Answered question

2022-09-26

An insect pest population doubles every 18 days. If an insecticide kills 90% of the insects, how often should it be applied to keep in insect population in check?
I tried using this exponential growth/decay model:
$A=P{\left(\frac{{A}_{1}}{{P}_{1}}\right)}^{t/{t}_{1}}$
where ${P}_{1}$ (initial value), ${A}_{1}$ (new value), t (time period).
however, might this be the right manner to do it? i'm no longer quite sure how to interpret (and for this reason clear up) this problem.

### Answer & Explanation

vidovitogv5

Beginner2022-09-27Added 10 answers

So you have an insect population that doubles every 18 days. Let's call it ${m}_{0}$. After 18 days, you will have 2${m}_{0}$, after another 18 days, you will have 4${m}_{0}$. The idea is that you model this growth with a function:
f(t)=something,
but we know some values of this function:

and so on.
We know the form of the function:

To find n, we do:

Our function is:

Remember that the insect population is always growing, and it will hit the desired number of insects (or a critical number of insects) at ${t}_{c}$ (critical time). Let's call that critical number of insects ${m}_{c}$. Then you apply the insecticide and now the insect population is:
$f\left({t}_{c}\right)={m}_{0}\cdot {2}^{{t}_{c}/18}=0.1{m}_{c}$
Then you have a new function of insect population:

Since the question is "how often should it be applied to keep the insect population in check?", the insect population will grow and it will hit the desired number of insects, mc after some time t, we want to know that time (in days):
$0.1{m}_{c}\cdot {2}^{t/18}={m}_{c},$
and the result is (solve for t!):
t=59.79 days.

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